The numbr of real roots of the equation `x ^(2013)+ e ^(20144x) =0` is

Answer

To find the number of real roots of the equation \( x^{2013} + e^{20144x} = 0 \), we can follow these steps:

### Step 1: Define the function
Let \( f(x) = x^{2013} + e^{20144x} \).

### Step 2: Analyze the function
We need to determine the behavior of \( f(x) \) to find the number of real roots.

### Step 3: Find the derivative
Calculate the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^{2013}) + \frac{d}{dx}(e^{20144x}) = 2013x^{2012} + 20144e^{20144x} \]

### Step 4: Analyze the derivative
Both terms in \( f'(x) \) are always positive:
- \( 2013x^{2012} \) is non-negative for all \( x \) since it is a power of \( x \) with an even exponent.
- \( 20144e^{20144x} \) is always positive since the exponential function is always positive for any real number \( x \).

Thus, \( f'(x) > 0 \) for all \( x \).

### Step 5: Determine the behavior of \( f(x) \)
Since \( f'(x) > 0 \), the function \( f(x) \) is strictly increasing. Now, we need to evaluate the limits of \( f(x) \) as \( x \) approaches negative and positive infinity.

1. As \( x \to -\infty \):
\[ f(x) = x^{2013} + e^{20144x} \to -\infty \quad (\text{since } e^{20144x} \to 0 \text{ and } x^{2013} \to -\infty) \]

2. As \( x \to +\infty \):
\[ f(x) = x^{2013} + e^{20144x} \to +\infty \quad (\text{both terms tend to } +\infty) \]

### Step 6: Conclusion about the roots
Since \( f(x) \) is strictly increasing, starts from \( -\infty \) as \( x \to -\infty \), and goes to \( +\infty \) as \( x \to +\infty \), it must cross the x-axis exactly once. Therefore, there is exactly one real root.

### Final Answer
The number of real roots of the equation \( x^{2013} + e^{20144x} = 0 \) is **1**.

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